Análise Taxa de Desemprego
raiz_unit(serie_desemprego)
## $ADF
##
## Augmented Dickey-Fuller Test
##
## data: ts
## Dickey-Fuller = -2.0429, Lag order = 6, p-value = 0.5579
## alternative hypothesis: stationary
##
##
## $PP
##
## Phillips-Perron Unit Root Test
##
## data: ts
## Dickey-Fuller Z(alpha) = -2.1946, Truncation lag parameter = 5, p-value
## = 0.963
## alternative hypothesis: stationary
##
##
## $KPSSL
##
## KPSS Test for Level Stationarity
##
## data: ts
## KPSS Level = 2.6349, Truncation lag parameter = 5, p-value = 0.01
##
##
## $Tabela
## Testes H0 p_valor Conclusao
## 1 Augmented Dickey-Fuller Tendencia 0.5579 Tendencia
## 2 Phillips-Perron Unit Root Tendencia 0.9630 Tendencia
## 3 KPSS Test for Level NAO tendencia 0.0100 Tendencia
tend_determ(serie_desemprego)
## $CS
##
## Cox Stuart test
##
## data: ts
## statistic = 25, n = 133, p-value = 1.718e-13
## alternative hypothesis: non randomness
##
##
## $CeST
##
## Cox and Stuart Trend test
##
## data: ts
## z = 9.4868, n = 270, p-value < 2.2e-16
## alternative hypothesis: monotonic trend
##
##
## $MannKT
##
## Mann-Kendall trend test
##
## data: ts
## z = -12.049, n = 270, p-value < 2.2e-16
## alternative hypothesis: true S is not equal to 0
## sample estimates:
## S varS tau
## -1.786200e+04 2.197416e+06 -4.965964e-01
##
##
## $MannK
## tau = -0.497, 2-sided pvalue =< 2.22e-16
##
## $KPSST
##
## KPSS Test for Trend Stationarity
##
## data: ts
## KPSS Trend = 0.83108, Truncation lag parameter = 5, p-value = 0.01
##
##
## $Tabela
## Testes H0 p_valor Conclusao
## 1 Cox Stuart NAO tendencia 0.00 Tendencia
## 2 Cox and Stuart Trend NAO tendencia 0.00 Tendencia
## 3 Mann-Kendall Trend NAO tendencia 0.00 Tendencia
## 4 Mann-Kendall NAO tendencia 0.00 Tendencia
## 5 KPSS Test for Trend NAO tendencia 0.01 Tendencia
sazonalidade(serie_desemprego)
## $KrusW
## Test used: Kruskall Wallis
##
## Test statistic: 0.21
## P-value: 1
##
## $Fried
## Test used: Friedman rank
##
## Test statistic: 22.94
## P-value: 0.01805006
##
## $Tabela
## Testes H0 p_valor Conclusao
## 1 Kruskall Wallis NAO Sazonal 1.0000 NAO Sazonal
## 2 Friedman rank NAO Sazonal 0.0181 Sazonal
serie_desemprego_part <- ts_split(serie_desemprego, sample.out = 12)
serie_desemprego_train <- serie_desemprego_part$train
serie_desemprego_test <- serie_desemprego_part$test
modelo1 <- auto.arima(serie_desemprego_train)
summary(modelo1)
## Series: serie_desemprego_train
## ARIMA(1,2,1)(0,0,2)[12]
##
## Coefficients:
## ar1 ma1 sma1 sma2
## -0.3353 -0.4303 -0.1283 -0.1507
## s.e. 0.0941 0.0940 0.0638 0.0619
##
## sigma^2 = 4.064e-07: log likelihood = 1521.71
## AIC=-3033.42 AICc=-3033.18 BIC=-3015.69
##
## Training set error measures:
## ME RMSE MAE MPE MAPE
## Training set -1.551009e-05 0.0006300452 0.0004884268 -0.008591897 0.6315899
## MASE ACF1
## Training set 0.07268784 0.002654839
modelo1_fc <- forecast::forecast(modelo1, h = 12)
coeftest(modelo1)
##
## z test of coefficients:
##
## Estimate Std. Error z value Pr(>|z|)
## ar1 -0.335297 0.094064 -3.5646 0.0003645 ***
## ma1 -0.430321 0.093992 -4.5783 4.688e-06 ***
## sma1 -0.128297 0.063773 -2.0118 0.0442456 *
## sma2 -0.150657 0.061931 -2.4327 0.0149877 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
forecast::accuracy(modelo1_fc, serie_desemprego_test)[,c(1:3, 5)]
## ME RMSE MAE MAPE
## Training set -1.551009e-05 0.0006300452 0.0004884268 0.6315899
## Test set 2.258501e-03 0.0025234201 0.0022585006 5.1714145
test_forecast(actual = serie_desemprego,
forecast.obj = modelo1_fc,
test = serie_desemprego_test)